Kinetic and Potential Energy Position in SHM – Rankers Physics
Topic: Oscillation
Subtopic: Energy in SHM

Kinetic and Potential Energy Position in SHM

A particle is performing simple harmonic motion of amplitude \(A\) about origin. Then position at which kinetic energy of particle is 8 times of its potential energy at that instant is:
\(x = \frac{A}{2\sqrt{2}}\)
\(x = \frac{A}{8}\)
\(x = \frac{A}{3}\)
\(x = \frac{A}{9}\)

Solution:

The given condition is \(\text{KE} = 8\text{PE}\). Substituting the expressions: \(\frac{1}{2}m\omega^2(A^2 - x^2) = 8\left(\frac{1}{2}m\omega^2 x^2\right)\), which simplifies to \(A^2 - x^2 = 8x^2\) or \(9x^2 = A^2\). Thus, \(x = \frac{A}{3}\).

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